Saudi Arabian Mathematical Competitions

(a) Assume that the origin of the coordinates system is at $A$. Let $r$ be the inradius of $\triangle ABC$, and $I$ the incenter. Then $I(r, r)$ and the line $d$ has equation
$$d: y - r = m(x - r)$$
that is $y = r + m(x - r)$. We get
$$P\left(\frac{r(m-1)}{m}, 0\right), \quad Q(0, r(m-1)) .$$

It follows
$$\begin{aligned} b \cdot \frac{PB}{PA} & + c \cdot \frac{QC}{QA} = b \cdot \frac{c - \frac{r(m-1)}{m}}{\frac{r(m-1)}{m}} + c \cdot \frac{b + r(m-1)}{-r(m-1)} \\ & = b \cdot \frac{m c - r(m-1)}{r(m-1)} + c \cdot \frac{b + r(m-1)}{-r(m-1)} \\ & = b \cdot \frac{m c}{r(m-1)} - c \cdot \frac{b}{r(m-1)} - (b + c) \\ & = \frac{b c}{r(m-1)}(m-1) - (b + c) = \frac{b c}{r} - (b + c) . \end{aligned}$$
But $r = \frac{2K}{a + b + c} = \frac{b c}{a + b + c}$, hence $\frac{b c}{r} = a + b + c$, and we get:
$$b \cdot \frac{PB}{PA} + c \cdot \frac{QC}{QA} = (a + b + c) - (b + c) = a .$$

(b) From the previous relation we have
$$\frac{b}{a} \cdot \frac{PB}{PA} + \frac{c}{a} \cdot \frac{QC}{QA} = 1,$$
so from Cauchy-Schwarz inequality it follows
$$\left[\left(\frac{PB}{PA}\right)^{2} + \left(\frac{QC}{QA}\right)^{2}\right]\left(\frac{b^{2}}{a^{2}} + \frac{c^{2}}{a^{2}}\right) \geq \left(\frac{b}{a} \cdot \frac{PB}{PA} + \frac{c}{a} \cdot \frac{QC}{QA}\right)^{2} = 1$$
hence
$$\left(\frac{PB}{PA}\right)^{2} + \left(\frac{QC}{QA}\right)^{2} \geq 1$$
The minimum value is $1$ and it is obtained for the line $d$ satisfying the property
$$\frac{PB}{PA} \cdot \frac{a}{b} = \frac{QC}{QA} \cdot \frac{a}{c} .$$

(a) (Abdullah Al-Saeed). The relation is equivalent to
$$b \cdot \frac{c - PA}{PA} + c \cdot \frac{b - QA}{QA} = a$$
that is
$$\begin{equation*} bc\left(\frac{1}{PA} + \frac{1}{QA}\right) = a + b + c \tag{1} \end{equation*}$$
We know that $K[ABC] = \frac{b c}{2} = s r$, where $s$ is the semiperimeter of triangle $ABC$, hence
$$r = \frac{b c}{a + b + c},$$
and by replacing in (1) we obtain
$$\begin{equation*} \frac{r}{PA} + \frac{r}{QA} = 1 \tag{2} \end{equation*}$$
From the similarity we have $\frac{r}{PA} = \frac{QI}{QP}$ and $\frac{r}{QA} = \frac{PI}{QP}$, and a relation (2) follows.

(b) From the previous relation we have
$$\frac{b}{a} \cdot \frac{PB}{PA} + \frac{c}{a} \cdot \frac{QC}{QA} = 1,$$
so from Cauchy-Schwarz inequality it follows
$$\left[\left(\frac{PB}{PA}\right)^{2} + \left(\frac{QC}{QA}\right)^{2}\right]\left(\frac{b^{2}}{a^{2}} + \frac{c^{2}}{a^{2}}\right) \geq \left(\frac{b}{a} \cdot \frac{PB}{PA} + \frac{c}{a} \cdot \frac{QC}{QA}\right)^{2} = 1$$
hence
$$\left(\frac{PB}{PA}\right)^{2} + \left(\frac{QC}{QA}\right)^{2} \geq 1$$
The minimum value is $1$ and it is obtained for the line $d$ satisfying the property
$$\frac{PB}{PA} \cdot \frac{a}{b} = \frac{QC}{QA} \cdot \frac{a}{c} .$$